1556

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
#define ll long long
using namespace std;
const int maxn=22;
const double pi=acos(-1.0);
const double D_MAX=1e100;
const double D_MIN=-1e100;
const double eps=1e-9;
int sgn(double x){ if(fabs(x) < eps)return 0;  if(x >0) return 1; return -1; }
int dcmp(double x, double y){ if(fabs(x - y) < eps) return 0; if(x > y) return 1;return -1;}
void usehanshu(){double x;}//floor(x)向下取整函数ceil(x)向上取整函数round(x)四舍五入函数
struct Point  { double x,y; Point(double x=0,double y=0):x(x),y(y) {};  };
struct Segment{ Point a,b;  Segment(Point x,Point y)     { a=x;b=y; };   };
struct Line   { Point a,b;  Line(Point x,Point y)        { a=x;b=y; };   };
typedef Point Vector;
Vector operator + (Vector A, Vector B){ return Vector(A.x+B.x, A.y+B.y); } // 向量相加
Vector operator - (Point  A, Point  B){ return Vector(A.x-B.x, A.y-B.y); } // 向量生成
double operator * (Vector A, Vector B){ return A.x*B.x-A.y*B.y;          } // 点积
double operator ^ (Vector A, Vector B){ return A.x*B.y-A.y*B.x;          } // 叉积
double Dot(Vector A, Vector B)   { return A.x*B.x + A.y*B.y; }  // 点积
double Cross(Vector A, Vector B) { return A.x*B.y-A.y*B.x;   }  // 叉积
double Length(Vector A) { return sqrt(Dot(A, A));}           // 向量长度
double Angle(Vector A, Vector B){ return acos(Dot(A, B)/Length(A)/Length(B));}  // 角度
double Area2(Point A, Point B, Point C) { return Cross(B-A, C-A); }  // 四边形面积
double dis(Point A,Point B) { return sqrt( (A.x-B.x)*(A.x-B.x)+(A.y-B.y)*(A.y-B.y) );  }
Vector Rotate(Vector A, double rad){ return Vector(A.x*cos(rad)-A.y*sin(rad), A.x*sin(rad)+A.y*cos(rad));}//rad为弧度 且为逆时针旋转的角
Vector Normal(Vector A)            {double L = Length(A);return Vector(-A.y/L, A.x/L);}//向量A左转90°的单位法向量
bool ToLeftTest(Point a, Point b, Point c){return Cross(b - a, c - b) > 0;}
bool xx(Line A,Point a){if( sgn( Cross(A.a-a,A.b-a) )==0 ) return 1;return 0;}  // 直线和点
void XX(Line A,Line B)  // 直线和直线的情况
{
    Point a=A.a; Point b=A.b;Point c=B.a; Point d=B.b;
    double A1=b.y-a.y,B1=-(b.x-a.x),C1=b.y*a.x-b.x*a.y;
    double A2=d.y-c.y,B2=-(d.x-c.x),C2=d.y*c.x-d.x*c.y;
    double k=A1*B2-A2*B1;
    if(fabs(k)<eps)
    {
        if( fabs( C2*A1-C1*A2)<eps && fabs(B1*C2-C1*B2)<eps ) printf("LINE
");
        else                   printf("NONE
");
    }
    else
    {
            double x=-(B1*C2-C1*B2)*1.000000000/k;
            double y=(A1*C2-C1*A2)*1.00000000/k;
            printf("POINT %.2f %.2f
",x,y);
    }
}

//   ------------------------------------------------ show time ----------------------------------------------------
//int a[maxn];
double dp[maxn][maxn];
vector<Point>   vs[maxn];
vector<Segment> vd[maxn];
bool make(Line a,Segment b)
{
    Point x=a.a;
    Point y=a.b;
    Point qq=b.a;
    Point pp=b.b;
    if( Cross(x-qq,y-qq )*Cross(x-pp,y-pp)>=0 ) return 1;
    return 0;
}
void check(int p)
{
    for(int i=0;i<vs[p].size();i++)  //  qiu de hang  shang de dian
    {
        dp[p][i]=D_MAX;
        Point X=vs[p][i];   // Point

        for(int j=0;j<p;j++)
        {
            for(int k=0;k<vs[j].size();k++)
            {
                Point Y=vs[j][k];  // Point
                int flag=1;
                for(int x=j+1;x<p;x++)    //  zhong jian hang
                {
                    for(int q=0;q<vd[x].size();q++)
                    {
                        Segment K=vd[x][q]; if(make(Line(X,Y),K)==0) {flag=0;break;}
                    }
                    if(flag==0) break;
                }
                if(flag==1)
                {

                    dp[p][i]=min(dp[p][i],dp[j][k]+dis(X,Y) );
                }
            }
        }
    }

}
int main()
{
    while(1)
    {
        int n; scanf("%d",&n);
        if(n==-1) break;
        vs[0].push_back(Point(0,5)); dp[0][0]=0;
        for(int i=1;i<=n;i++)
        {
            double x,a,b,c,d;
            scanf("%lf %lf %lf %lf %lf",&x,&a,&b,&c,&d);
            vs[i].push_back(Point(x,a));
            vs[i].push_back(Point(x,b));
            vs[i].push_back(Point(x,c));
            vs[i].push_back(Point(x,d));  // 4 ge dian
            vd[i].push_back(Segment( Point(x,0.0),Point(x,a))  );
            vd[i].push_back(Segment( Point(x,b),Point(x,c))    );
            vd[i].push_back(Segment( Point(x,d),Point(x,10.0)) );
            check(i);
        }
        vs[n+1].push_back(Point(10,5));
        check(n+1);
        printf("%.2f
",dp[n+1][0]);

        for(int i=0;i<=n+1;i++){vs[i].clear(); vd[i].clear();}
        for(int i=0;i<=n+1;i++)
            for(int j=0;j<=5;j++)
            dp[i][j]=0;
    }

}